Dirac distribution verifies $x\delta(x)=0$ Let $\phi:\mathbb R \to \mathbb R$ be a test function. We denote $D(\mathbb R)$ the set of test functions. The dirac distribution 
$$\delta :D(\mathbb R)\to \mathbb R$$ is defined by:
$$<\delta , \phi>=\phi(0)$$ Now take any smooth function $g:\mathbb R\to \mathbb R$, and define the product distribution $g.\delta$ by:
$$<g.\delta,\phi>=<\delta,g\phi>$$
From this definition we see that 
$$<g.\delta,\phi>=g(0)\phi(0)$$
Now i'm reading the following problem prove that $x\delta(x)=0$ and $x\delta'(x)=-\delta(x)$ wihtout any firther information, and I want to understand the meaning of these formulas regarding the definitions above, I mean what is $x$ and what is $\delta(x)$ and what is the product $x\delta(x)$? It seems like he takes $g$ for being the identity $id :\mathbb R\to \mathbb R$ but this would give 
$$<id.\delta, \phi>=id(0)\phi(0)=0*\phi(0)=0$$ but i don't see where $x\delta(x)=0$ come from. thank you for your help!
 A: By definition we can write for any test function $\phi$
$$\langle x\delta_0,\phi(x)\rangle = \langle  \delta_0, x\phi(x)\rangle = 0\cdot \phi(0)=0=\langle 0,\phi(x)\rangle,$$
which means that $x\delta_0=0$ in the sense of distributions.
In the same spirit, we write
$$\langle x\delta'_0,\phi(x)\rangle = \langle  \delta_0', x\phi(x)\rangle =- \langle \delta_0,(x\phi(x))'\rangle = -\langle \delta_0,\phi(x)+x\phi'(x) \rangle = -\phi(0)-0\cdot\phi'(0) = -\phi(0) = -\langle \delta_0, \phi(x) \rangle,$$
which means that $x\delta'_0= -\delta_0$ in the sense of distributions.
A: *

*For a distribution $u$ and polynomials $P,Q$, we define the distribution $P(M_{{\rm id}_{\mathbb{R}}}) Q(\partial)u$ as
$$\tag{1} \left(P(M_{{\rm id}_{\mathbb{R}}}) Q(\partial)u\right) [\varphi]~:=~ u\left[ Q(-\partial)P(M_{{\rm id}_{\mathbb{R}}})\varphi\right], $$
where $\varphi$ is a test function;
where $M_f(g):=fg$ denotes the multiplication operator with the function $f$;
where $\partial(f)=f^{\prime}$ is differentiation;
and where ${\rm id}_{\mathbb{R}}$ is the identity function: $\mathbb{R}\to \mathbb{R}$. Note that the multiplication operator $M_{{\rm id}_{\mathbb{R}}}$ is often denoted by the indeterminate $x$ by a slight abuse of notation. 

*The Dirac delta distribution $u=\delta$ is defined as 
$$\tag{2}\delta [\varphi]~:=~\varphi(0).$$

*The distribution identities 
$$\tag{3} M_{{\rm id}_{\mathbb{R}}}\delta~=~0\qquad\text{and}\qquad M_{{\rm id}_{\mathbb{R}}}\partial\delta~=~-\delta,$$ 
(which is often written as 
$$\tag{4} x\delta(x)~=~0\qquad\text{and}\qquad x\delta^{\prime}(x)~=~-\delta(x),$$ 
with a slight abuse of notation), follow from 
$$\tag{5} \left(M_{{\rm id}_{\mathbb{R}}}\delta\right)[\varphi]~\stackrel{(1)}{=}~\delta\left[M_{{\rm id}_{\mathbb{R}}} \varphi\right]~\stackrel{(2)}{=}~0\varphi(0)~=~0~=~0[\varphi],$$ 
and 
$$ \left(M_{{\rm id}_{\mathbb{R}}}\partial\delta\right)[\varphi]~\stackrel{(1)}{=}~\delta\left[-\partial M_{{\rm id}_{\mathbb{R}}} \varphi\right]~\stackrel{\text{Leibniz' rule}}{=}~\delta\left[-\varphi- M_{{\rm id}_{\mathbb{R}}} \partial\varphi\right]$$ 
$$\tag{6}~\stackrel{(2)}{=}~-\varphi(0)-0\varphi^{\prime}(0)~\stackrel{(2)}{=}~-\delta[\varphi],$$
respectively.
